Given the function $f: X \rightarrow \mathbb{R}$ with $\lim_{x\rightarrow a}f(x) = L$. Prove that $\lim_{x\rightarrow a}|f(x)| = |L|$.

$\forall \epsilon > 0 \exists\delta>0 s.t. 0<|x-a|< \delta \Rightarrow ||f(x)| - |L||\leq |f(x) - L| < \epsilon$.

I'm not really sure regarding this step $$||f(x)| - |L||\leq |f(x) - L| < \epsilon$$

Is it correct?

  • $\begingroup$ When you write $||f(x) - |L||$ do you mean $||f(x)| - |L||$ or do you genuinely think it's the former? $\endgroup$ – Zain Patel May 10 '17 at 3:08
  • $\begingroup$ @ZainPatel, it was a typo. I meant $||f(x)| -|L||$. $\endgroup$ – KirkLand May 10 '17 at 22:18

you have it nearly correct. the inequality you need is (with $a=f(x)$ and $b=L$) $$ ||a|-|b|| \le |a-b| $$ which holds in fact for any complex numbers $a,b$. it states that the minimum distance between two concentric circles is no greater that the distance separating any two points, one on each circle.

  • $\begingroup$ Thank you for that visualisation. $\endgroup$ – Calvin Khor May 10 '17 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.